Solve the following problem : In the following probability distribution of a r.v.X. x 1 2 3 4 5 P (x) 120 320 a 2a 120 Find a and obtain the c.d.f. of X. - Mathematics and Statistics

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Sum

Solve the following problem :

In the following probability distribution of a r.v.X.

x 1 2 3 4 5
P (x) `(1)/(20)` `(3)/(20)` a 2a `(1)/(20)`

Find a and obtain the c.d.f. of X.

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Solution

Since the given table represents a p.m.f. of r.v. X,

\[\sum\limits_{x=1}^{5}\text{P}(x) = 1\]

∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

∴ `(1)/(20) + (3)/(20) + "a" + 2"a" + (1)/(20)` = 1

∴ 3a = `1 - (5)/(20)`

∴ 3a = `1 - (1)/(4)`

∴ 3a = `(3)/(4)`

∴ a = `(1)/(4)`
By definition of c.d.f.,
F(x) = P(X ≤ x)

F(1) = P(X ≤ 1) = P(1) = `(1)/(20)`

F(2) = P(X ≤ 2) = F(1) + P(2)

= `(1)/(20) + (3)/(20) = (4)/(20)`

F(3) = P(X ≤ 3)
= F(2) + P(3)

= `(4)/(20) + "a" = (4)/(20) + (1)/(4) = (4)/(20) + (5)/(20) = (9)/(20)`

F(4) = P(X ≤ 4)
= F(3) + P(4)

= `(9)/(20) + 2"a" = (9)/(20) + (1)/(2) = (9)/(20) + (10)/(20) = (19)/(20)`

F(5) = P(X ≤ 5)
= F(4) + P(5)

= `(19)/(20) + (1)/(20)` = 1
∴ c.d.f. of X is as follows:

xi 1 2 3 4 5
F(xi) `(1)/(20)` `(4)/(20)` `(9)/(20)` `(19)/(20)` 1
Concept: Probability Distribution of a Continuous Random Variable
  Is there an error in this question or solution?
Chapter 8: Probability Distributions - Part I [Page 155]

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